Energy of AP and AR
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Energy of AP and AR
I wonder how the energy of AP and AR is computed. It is always said that the energy is limited by law at 7.5J but when I compute the energy from the muzzle velocity I have measured and the mass of the pellet, I get something much higher.
Example:
- for my LP10 I measured a velocity of 172m/s with a RWS R10 0.45g pellet. So the power is 172**2 * 0.00045 = 13.31J.
- for my AR Anschtuz 8002, also supposed to comply with 7.5J rule, it is even worse since the muzzle velocity I measured is 195m/s, so its energy is 17.11J.
These two guns have their muzzle velocity adjustement factory set. I didn't modify it.
What's wrong here, is the velcocity to use to compute the official energy different from the muzzle velocity ? I must say that the muzzle velocity I measured are not real muzzle velocity since they are measured at about 2m from the muzzle.
Example:
- for my LP10 I measured a velocity of 172m/s with a RWS R10 0.45g pellet. So the power is 172**2 * 0.00045 = 13.31J.
- for my AR Anschtuz 8002, also supposed to comply with 7.5J rule, it is even worse since the muzzle velocity I measured is 195m/s, so its energy is 17.11J.
These two guns have their muzzle velocity adjustement factory set. I didn't modify it.
What's wrong here, is the velcocity to use to compute the official energy different from the muzzle velocity ? I must say that the muzzle velocity I measured are not real muzzle velocity since they are measured at about 2m from the muzzle.
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well, the 6.7 is just because the equation includes the 1/2
1/2mv^2 i think is the correct equation for energy, but this still puts his rifle above the limit at about 8.6J.
I think this is the main problem with the calculations, the missing 1/2.
I don't know why the rifle is above the 7.5 and that could be due to the things mentioned in the post, about the law's measurements and where they take them (at the endpoint, at the muzzle, etc) and varying factors like that.
1/2mv^2 i think is the correct equation for energy, but this still puts his rifle above the limit at about 8.6J.
I think this is the main problem with the calculations, the missing 1/2.
I don't know why the rifle is above the 7.5 and that could be due to the things mentioned in the post, about the law's measurements and where they take them (at the endpoint, at the muzzle, etc) and varying factors like that.
E=0.5*m*v^2
E = 0.5 * m * v ^2
where
E = energy [Joules]
m = mass [kg]
v = velocity [m/s]
and "^2" denotes square (i.e. v^2 = v*v)
So with a pellet of 0.53 grams = 0.00053 kg, (e.g. RWS R10)
and a velocity of 170 m/s,
you get E = 7.66 joules: just a little bit above the famous 7.5.
But where to measure the v?
I found a document that mentions the speed at 2.5 m from the muzzle should be used according to Belgian law:
http://www.schietsport.be/scriptie_nieuwe_wapenwet.pdf
but apparently with an exception for rifles that "can only be used for sports".
F.w.i.w.: at 2.5 m from the muzzle, velocity will be down from V0=170 to 165 m/s, so E will be 6.8 J then.
Having said all that: I do not know WHICH laws (i.e. of which countries) specifiy a 7.5 J limit. Dutch law no longer has an energy limit for air rifles.
where
E = energy [Joules]
m = mass [kg]
v = velocity [m/s]
and "^2" denotes square (i.e. v^2 = v*v)
So with a pellet of 0.53 grams = 0.00053 kg, (e.g. RWS R10)
and a velocity of 170 m/s,
you get E = 7.66 joules: just a little bit above the famous 7.5.
But where to measure the v?
I found a document that mentions the speed at 2.5 m from the muzzle should be used according to Belgian law:
http://www.schietsport.be/scriptie_nieuwe_wapenwet.pdf
but apparently with an exception for rifles that "can only be used for sports".
F.w.i.w.: at 2.5 m from the muzzle, velocity will be down from V0=170 to 165 m/s, so E will be 6.8 J then.
Having said all that: I do not know WHICH laws (i.e. of which countries) specifiy a 7.5 J limit. Dutch law no longer has an energy limit for air rifles.
Last edited by JeroenH on Thu Nov 15, 2007 1:17 pm, edited 3 times in total.
Re: Energy of AP and AR
What is wrong is that you are calculating "halv-momentum" (mass *velocity*0.5). That calculation has less meaning I think, and is not an expression for energy, btw, and is not measured in Joules.I wonder how the energy of AP and AR is computed.
Example:
- for my LP10 I measured a velocity of 172m/s with a RWS R10 0.45g pellet. So the power is 172**2 * 0.00045 = 13.31J.
What's wrong here, is the velcocity to use to compute the official energy different from the muzzle velocity ?
Translatory (kinetic) energy is 0.5*mass*velocity*velocity, and is often measured in Newtonmeters pr. second, or Joules (J).
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Re: E=0.5*m*v^2
In the UK the limits are 6 foot pounds muzzle energy for air pistol and 12 foot pounds muzzle energy for air rifle before you need a firearm certificate.JeroenH wrote:Having said all that: I do not know WHICH laws (i.e. of which countries) specifiy a 7.5 J limit. Dutch law no longer has an energy limit for air rifles.
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Power Calculations
Perhaps this will help.
I can't remember where it came from but it works.
To find out the Energy of a rifle = fps x fps x grains / 450240 = ft/lbs.
To convert ft/lbs to Joules = ft/lbs x 1.355817 = Joules.
To convert Joules to ft/lbs = Joules x 0.737562149 = ft/lbs.
1 grain = 0.06479891 grams = 0.002285714 oz (ounces)
1 Kilogram = 2.2046 lbs (pounds)
1 inch = 2.54 centimeters
.177 caliber is 4.5mm .20 caliber is 5.05mm .22 caliber is 5.5mm .25 caliber is 6.35mm
There is a programme called Agpower which I can not get to zip so that I can post it here. It's very useful in that you give it a pellet calibre and name and it will tell you what MV or what power in FT/LB provided you have one or the other
Using it I get the RWS Hobby .177 at 6.9grains to give you 4.91 ft/lb at the muzzle. This is well inside UK regs for a pistol so no problem at all legally.
The RWS Field point at 7.7Grains is 5.44 ft/lb and the RWS Super Hollow Point at 7.4grains is 5.2 ft/lb.
These were the closest I could find to .45 grammes and calculated at 564 ft/sec.
Hope this helps.
Target Bunny
I can't remember where it came from but it works.
To find out the Energy of a rifle = fps x fps x grains / 450240 = ft/lbs.
To convert ft/lbs to Joules = ft/lbs x 1.355817 = Joules.
To convert Joules to ft/lbs = Joules x 0.737562149 = ft/lbs.
1 grain = 0.06479891 grams = 0.002285714 oz (ounces)
1 Kilogram = 2.2046 lbs (pounds)
1 inch = 2.54 centimeters
.177 caliber is 4.5mm .20 caliber is 5.05mm .22 caliber is 5.5mm .25 caliber is 6.35mm
There is a programme called Agpower which I can not get to zip so that I can post it here. It's very useful in that you give it a pellet calibre and name and it will tell you what MV or what power in FT/LB provided you have one or the other
Using it I get the RWS Hobby .177 at 6.9grains to give you 4.91 ft/lb at the muzzle. This is well inside UK regs for a pistol so no problem at all legally.
The RWS Field point at 7.7Grains is 5.44 ft/lb and the RWS Super Hollow Point at 7.4grains is 5.2 ft/lb.
These were the closest I could find to .45 grammes and calculated at 564 ft/sec.
Hope this helps.
Target Bunny
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Hi Again
With a bit of luck this is the programme that will calculate velocity or muzzle energy.
With a bit of luck this is the programme that will calculate velocity or muzzle energy.
- Attachments
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- agpower.zip
- If I'm lucky this will unzip and install to let you calculate power or speed.
- (273.81 KiB) Downloaded 335 times
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Hi Again, again
This is a better calculator but you need to get it direct from Airhog at http://www.airhog.com/convert.htm.
It does Metric and Imperial at the same time.
Regards
Jim
This is a better calculator but you need to get it direct from Airhog at http://www.airhog.com/convert.htm.
It does Metric and Imperial at the same time.
Regards
Jim